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3 years ago

What is the solution to the equation A/2= -5

Mathematics

2 answers:

dem82 [27]3 years ago

5 0

Answer:

-10

Step-by-step explanation:

It is the easiest equation.

A/2= -5

At first, we have to multiply both the sides by 2. Therefore, we can get,

$\frac{A*2}{2}$ = (-5 × 2)

or, A = -10

Therefore, the value of A is -10. It remains negative because we cannot multiply both the sides by -1. If we do that, we cannot determine the constant.

Answer: A = -10

murzikaleks [220]3 years ago

3 0

Answer:

A = -10

Step-by-step explanation:

A/2 = -5

Multiply both sides by the denominator of the fraction

We have A/2 x2 = -5 x 2

A = -10

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What is the value of n in the equation 1/2(n – 4) – 3 = 3 – (2n + 3)?

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(1/2)(n - 4) - 3 = 3 - (2n + 3
Distribute the (1/2) on the left and the -1 on the right.
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3 years ago

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Apply Crammer's Rule to find the solution to the following quations .2x + 3y = 1, 3x + y = 5

Bond [772]

Answer:

The solution to the equation system given is:

<u>x = 2</u>
<u>y = -1</u>

Step-by-step explanation:

First, we must know the equations given:

2x + 3y = 1
3x + y = 5

Following Crammer's Rule, we have the matrix form:

$\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] =\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]$

Now we solve using the determinants:

$x=\frac{\left[\begin{array}{ccc}1&3\\5&1\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(1*1)-(5*3)}{(2*1)-(3*3)} = \frac{1-15}{2-9} =\frac{-14}{-7} = 2$

$y=\frac{\left[\begin{array}{ccc}2&1\\3&5\end{array}\right]}{\left[\begin{array}{ccc}2&3\\3&1\end{array}\right] } =\frac{(2*5)-(3*1)}{(2*1)-(3*3)}=\frac{10-3}{2-9} =\frac{7}{-7}=-1$

Now, we can find the answer which is x= 2 and y= -1, we can replace these values in the equation to confirm the results are right, with the first equation:

2x + 3y = 1
2(2) + 3(-1)= 1
4 - 3 = 1
1 = 1

And, with the second equation:

3x + y = 5
3(2) + (-1) = 5
6 - 1 = 5
5 = 5

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3 years ago